# Confused about chain rule in Frechet derivative

I've recently started to learn about Frechet derivatives and now have a simple example which I'm not sure if I've solved correctly. To be honest, I've only got a poor understanding of how it works so please assume I am starting at zero. Given $$\delta = \sum_i \lambda_iX_i$$ for some matrices $$X_i$$ and real numbers $$\lambda_i$$ and $$N(X)$$ being a linear transformation on $$X$$, find the Frechet derivative

$$\frac{\partial}{\partial \lambda_l}Tr(A\log(N(\delta)),$$

for some $$l$$.

Redoing my attempted solution after reading greg's answer. Would appreciate if anyone can comment on the correctness of it!

We have \begin{align} d Tr(A\log(N(\delta)) &= Tr(dA\log(N(\delta)) \\ &= Tr\left(\left(\int_0^{\infty}\frac{1}{1+tN(\delta)}A\mathrm{d}t\frac{1}{1+tN(\delta)}\right)dN(\delta)\right) \\ &= Tr\left(\left(\int_0^{\infty}\frac{1}{1+tN(\delta)}A\mathrm{d}t\frac{1}{1+tN(\delta)}\right)N(d\delta)\right) \\ &= Tr\left(\left(\int_0^{\infty}\frac{1}{1+tN(\delta)}A\mathrm{d}t\frac{1}{1+tN(\delta)}\right)N(X_k d\lambda_k)\right) \\ \end{align}

I have used the fact that $$Tr$$ and $$N()$$ are both linear operators and the solution posted here to write the function $$d\ (A\log X)$$. I believe that this cannot proceed further unless I can say something about $$N()$$ to rewrite $$N(X_k d\lambda_k)$$ as $$M(X_k) d\lambda_k$$. If and only if I could do that step, then I have

\begin{align} d Tr(A\log(N(\delta)) &= Tr\left(\left(\int_0^{\infty}\frac{1}{1+tN(\delta)}A\mathrm{d}t\frac{1}{1+tN(\delta)}\right)M(X_k) d\lambda_k\right)\\ \frac{\partial}{\partial\lambda_k}Tr(A\log(N(\delta)) &= \left(\left(\int_0^{\infty}\frac{1}{1+tN(\delta)}A\mathrm{d}t\frac{1}{1+tN(\delta)}\right)M(X_k)\right)^T \end{align}

Define the matrix \eqalign{ Y &= \sum_{i=k}^N \lambda_kX_k = \lambda_kX_k \cr } where the expression on the far RHS uses the index summation convention.
Now calculate the differential and derivative for the trace of a simple nonlinear function. \eqalign{ \phi &= {\rm Tr}\Big(Y^3\Big) \cr d\phi &= {\rm Tr}\Big(3Y^2\,dY\Big) = {\rm Tr}\Big(3Y^2X_k\,d\lambda_k\Big) \cr \frac{d\phi}{d\lambda_k} &= {\rm Tr}\Big(3Y^2X_k\Big) \cr } A general function $$f(Y)$$ follows the same pattern. \eqalign{ \psi &= {\rm Tr}\Big(f(Y)\Big) \cr \frac{d\psi}{d\lambda_k} &= {\rm Tr}\Big(f'(Y)\,X_k\Big) \cr } where $$f'$$ denotes the ordinary derivative of the function.
• Thank you for this answer. Using this and your comments on my other question, I've redone the solution. I believe it can only proceed if I make an assumption about $N()$, as I have clarified in the edited question. If you wouldn't mind taking a moment to read it, can I check that this is the correct solution to the original problem I posed? – user1936752 Mar 17 at 22:23